• Option 1

  *This doesn't seem like a great idea to me, because it complicates the SQL to select all properties applied to a feature…

  It does not necessarily complicate query SQL (see conclusion below).

  …and doesn't readily scale to more conditions…

  It scales readily to more conditions, as long as there are still a fixed number of conditions, and there aren't dozens or hundreds.

  However, it does enforce the requirement of a certain number of conditions per (X,Y) pair. In fact, it is the only option here that does so.*

  It does, and although you say in a comment that this is "the least important of my requirements", you haven't said it doesn't matter at all.

• Option 2

  One downside to this is that it doesn't specify the number of conditions for each pair. Another is that when I am only considering the initial relationship…I then have to add a DISTINCT clause to avoid duplicate entries…

  I think you can dismiss this option because of the complications you mention. The objx_objy table is likely to be the driving table for some of your queries (eg "select all properties applied to a feature", which I am taking to mean all properties applied to an objx or objy). You can use a view to pre-apply the DISTINCT so it is not a matter of complicating queries, but that's going to scale very badly performance-wise for very little gain.

• Option 3

  Does it make sense though to create a new ID that identifies nothing other than existing IDs?

  No, it doesn't — Option 4 is better in every regard.

• Option 4

  …it basically duplicates an entire table multiple times (or feels that way, anyway) so also doesn't seem ideal.

  This option is just fine — it is the obvious way of setting up the relations if the number of properties is variable or subject to change

Conclusion

My preference would be option 1 if the number of properties per objx_objy is likely to be stable, and if you can't imagine ever adding more than a handful extra. It is also the only option that enforces the 'number of properties = 3' constraint — enforcing a similar constraint on option 4 would likely involve adding c1_p_id… columns to the xy table anyway*.

If you really don't care much about that condition, and you also have reason to doubt that the number of properties condition is going to be stable then choose option 4.

If you aren't sure which, choose option 1 — it is simpler and that is definitely better if you have the option, as others have said. If you are put off option 1 "…because it complicates the SQL to select all properties applied to a feature…" I suggest the creating a view to provide the same data as the extra table in option 4:

option 1 tables:

create table prop(id integer primary key);
create table objx(id integer primary key);
create table objy(id integer primary key);

create table objx_objy(
  x_id integer references objx
, y_id integer references objy
, c1_p_id integer not null references prop
, c2_p_id integer not null references prop
, c3_p_id integer not null references prop
, primary key (x_id, y_id)
);

insert into prop(id) select generate_series(90,99);
insert into objx(id) select generate_series(10,12);
insert into objy(id) select generate_series(20,22);

insert into objx_objy(x_id,y_id,c1_p_id,c2_p_id,c3_p_id)
select objx.id, objy.id, 90, 91, 90+floor(random()*10)
from objx cross join objy;

view to 'emulate' option 4:

create view objx_objy_prop as
select x_id
     , y_id
     , unnest(array[1,2,3]) c_id
     , unnest(array[c1_p_id,c2_p_id,c3_p_id]) p_id
from objx_objy;

"select all properties applied to a feature":

select distinct p_id from objx_objy_prop where x_id=10 order by p_id;

/*
|p_id|
|---:|
|  90|
|  91|
|  97|
|  98|
*/

dbfiddle here

This looks like a straightforward one-to-many relationship, which as you suggested you can model by setting a foreign key in the child items table referring to the parent items table. As for performance, you won't have any kind of performance hit at all for a small site. Modelling your schema properly is your priority.

Consider renaming the tables too. For example, if the shop offers a cleaning service then you could think of the tables in terms of Services (e.g. car) and Costs (seats, trunk, etc)